limit of sequence

Brainwave said:
Prove that lim [(3n + 1)/(7n - 4)] = 3/7 where n -> infinity
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You did this sort of thing back in algebra when you found horizontal asymptotes. You were taught various techniques back in calculus when you studies limits. You have been taught definitions, etc, in your current advanced course. Where are you stuck?

Please reply showing your work and reasoning so far. When you reply, please include any theorems, etc, that you believe you are expected to apply. Thank you! ;)
 
Theorem

I think this may be of little help be cause I have applied it. A sequence {a} is said to converge to a real number l, if for any real number E> 0. There exist a positive integer n(E) such that I a - l I < E for all n > n(E).
 
Yes, so you need to start by looking at ∣an−l∣\displaystyle |a_n- l| which, here, is ∣3n+17n−4−37∣\displaystyle \left|\frac{3n+1}{7n-4}- \frac{3}{7}\right|. Start by combining those two fractions.
 
We are asked to apply the theorem provided above not by the basic one. Thanks
 
Brainwave said:
We are asked to apply the theorem provided above not by the basic one. Thanks
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Why are you rating your threads 5 stars? And to your reply, you have provided no theorem, you only gave the definition for the limit of a sequence (when it exists). The replies you recieved above will help you find your n(ϵ)\displaystyle n(\epsilon)
 
Ok let me bring a typical example. Prove that lim (n/(n+1)) = 1 as n -> infinity. Solution. From the definition above, Let E > 0 be given we need to produce n(E) > 0 such that i n/(n+1) - 1 i < E for all n > n(E). We compute as follows, I n/(n+1) - 1 i = I -1/(n+1) I = 1/(n+1) < 1/n. We can se 1/n < E to get n > 1/E. Take n(E) = {1/E + 1}. Then we have i n/(n+1) - 1 I < E for all n>n(E).
 
Brainwave said:
Ok let me bring a typical example....
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Okay. But what is the point of this example? It merely supports the replies that you have received, indicating how you need to proceed.

Please show your work on this exercise, clearly stating where you are getting stuck in the process. Thank you. ;)
 
$|\frac{3n+1}{7n-4}-\frac{3}{7}|$=$\frac{19}{7(7n-4)}<E$ when $n>\frac{19+28E}{49E}$
 
Brainwave said:
My workings is[sic] attached below
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In future, kindly please consider typing out your text, or at least posting an image that isn't sideways or upside-down. Thank you.

For other viewers, the following is the text in the image:


Prove that lim⁡n → ∞ 3n + 17n − 4 = 37\displaystyle \displaystyle{\lim_{n\, \rightarrow \,\infty}}\, \dfrac{3n\, +\, 1}{7n\, -\, 4}\, =\, \dfrac{3}{7}

From the definition of a convergent sequence. Let ϵ > 0\displaystyle \epsilon\, >\, 0, we need to produce n(e) > 0\displaystyle n(e)\, >\, 0 such that:

. . . . .∣3n + 17n − 4∣ < ϵ, ∀ n > n(e)\displaystyle \left|\dfrac{3n\, +\, 1}{7n\, -\, 4}\right|\, <\, \epsilon,\, \forall\, n\, >\, n(e)

Therefore:

. . . . .∣21n + 7 − 21n + 127(7n − 4)∣ = ∣1949A − 4∣\displaystyle \left|\dfrac{21n\, +\, 7\, -\, 21n\, +\, 12}{7(7n\, -\, 4)}\right|\, =\, \left|\dfrac{19}{49A\, -\, 4}\right|


What is your n(e)\displaystyle n(e)? What do you mean by "therefore"? What is "A"?
 
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